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ad-work [718]
2 years ago
9

Studying this brochure from nasa, which explains more detail the instruments carried by the Juno spacecraft which scientific act

ivities will jump conduct on its trip to Jupiter? Check all that apply
Physics
1 answer:
adoni [48]2 years ago
3 0

Answer:

Juno scientific payload includes:

  • A gravity/radio science system (Gravity Science)
  • A six-wavelength microwave radiometer for atmospheric sounding and composition (MWR)
  • A vector magnetometer (MAG)
  • Plasma and energetic particle detectors (JADE and JEDI)
  • A radio/plasma wave experiment (Waves)
  • An ultraviolet imager/spectrometer (UVS)
  • An infrared imager/spectrometer (JIRAM)

Explanation:

Each mission of NASA has a specific set of instruments that it uses to perform scientific experiments on the desired heavenly body. In case of Juno, the mission for Jupiter has a series of instruments that would study domains of gravitational forces, magnetic effect, particle detection, radiation detection, UV/IR imaging, and plasma experiments.

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A ball is launched with initial speed v from ground level up a frictionless slope (This means the ball slides up the slope witho
amid [387]

Answer:

hmax = 1/2 · v²/g

Explanation:

Hi there!

Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.

KE = PE

Where KE is the initial kinetic energy and PE is the final potential energy.

The kinetic energy of the ball is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the ball

v = velocity.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the ball.

g = acceleration due to gravity (known value: 9.81 m/s²).

h = height.

At  the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:

PE = KE

m · g · hmax = 1/2 · m · v²

Solving  for hmax:

hmax = 1/2 · v² / g

4 0
3 years ago
Why does the moon orbit the earth as the earth orbits the sun?<br><br>will give brainliest​
AysviL [449]
As the Earth rotates, it also moves, or revolves, around the Sun. ... As the Earth orbits the Sun, the Moon orbits the Earth. The Moon's orbit lasts 27 1/2 days, but because the Earth keeps moving, it takes the Moon two extra days, 29 1/2, to come back to the same place in our sky.
6 0
3 years ago
A 5kg bag falls a verticle height of 10m before hitting the ground.
g100num [7]

Answer:

u = 7m {s}^{ - 1}

Explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

me(1) = me(2)

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

m \times g \times h   =   \frac{1}{2}  \times m \times u {}^{2} \\

we divide both sides of the equation with mass to make the math easier.

9.8 \times 10 =  \frac{1}{2}  \times u {}^{2}  \\  \frac{98}{2}  = u {}^{2}  \\ u { }^{2} = 49 \\ u = 7

7 0
3 years ago
An object dropped on Planet P falls 144 m in 6 seconds. What is the gravitational acceleration of Planet P ? Gravitational accel
Tju [1.3M]

Answer:

The gravitational acceleration of the planet is, g = 8 m/s²

Explanation:

Given data,

The distance the object falls, s = 144 m

The time taken by the object is, t = 6 s

Using the III equations of motion

                  S = ut + ½ gt²

∴                 g = 2S/t²

Substituting the given values,

                   g = 2 x 144 /6²

                      = 8 m/s²

Hence, the gravitational acceleration of the planet is, g = 8 m/s²

7 0
3 years ago
A plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. F
vaieri [72.5K]

While plane is moving under tailwind condition it took time "t"

so here we will have

t = \frac{d}{v_{net}}

here net speed of the plane will be given as

v_{net} = v + v_w

t = \frac{1000}{205 + v_w}

similarly when it moves under the condition of headwind its net speed is given as

v_{net} = v - v_w

now time taken to cover the distance is 2 hours more

t + 2 = \frac{1000}{205 - v_w}

now solving two equations

\frac{1000}{205 + v_w} + 2 = \frac{1000}{205 - v_w}

solving above for v_w we got

v_w = 40.4 mph

6 0
3 years ago
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